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When $R$ is a ring (not necessarily commutative, and not necessarily with unity), I have a result that tells me that for $x \in R$, the ideal generated by $x$, $(x) $, is $= I_{x} = \langle RxR \rangle + Rx + xR + \mathbb{Z}x$.

If I define $(y)$ in a similar way, namely that $(y) = I_{y} = \langle RyR \rangle + Ry + yR + \mathbb{Z}y$, then I have another result that says that the product of two ideals $IJ = \sum_{i=1}^{n}x_{i}y_{i}: \, x_{i} \in I, \, y_{i} \in J $.

I would like to use this result to put together a proof similar to the one Qyburn did in his question here, but in the case where $R$ is commutative, but is not necessarily a ring with unity.

The only problem is, once I have $\displaystyle (x)(y) = I_{x}\cdot I_{y} = \left\{\sum_{i=1}^{n}x_{i}y_{i}: x_{i} \in I_{x}, \, y_{i} \in I_{y}\right \} $, I'm not sure what the elements of each ideal should look like when I put them into the sum.

I.e., in Qyburn's proof, in the case where $R$ is a ring with unity, the principal ideals $(x) = I_{x} = \{rx: r \in R\}$, and are thus very easy to work with algebraically.

He goes on to say that $$(x)\cdot (y) = I_x \cdot I_y = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I_x, y_i \in I_y \biggr\} = \biggl\{\sum_{i=1}^n (r_i x) (s_i y) : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{x y \sum_{i=1}^n r_i s_i : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{x y r : r \in R \biggr\} = I_{xy} = (xy)\, .$$

In my case, $$ \displaystyle (x)(y) = I_{x} \cdot I_{y} = \left \{ \sum_{i=1}^{n} x_{i}y_{i}: \, x_{i} \in I_{x}, \, y_{i} \in I_{y} \right \} = \sum_{i=1}^{n} \left\{\left(\langle r_{i}xr_{i} \rangle + r_{i}x+xr_{i} + \mathbb{Z}x\right)\left(\langle s_{i}ys_{i} \rangle + s_{i}y+ys_{i} + \mathbb{Z}y \right): \, r_{i}, s_{i} \in R \right\}$$.

Is this correct? I guess the problem is is that I don't know how to manipulate the $\langle RxR \rangle$ and $\langle RyR \rangle$'s and the $\mathbb{Z}x$ and $\mathbb{Z}y$'s (what do these things even look like written out??)in order to get something to come out like Qyburn was able to.

If someone could please explain to me how to write out the algebraic details of these things correctly so that I could make the proof work, I would be most appreciative!

Thank you ahead of time.

user26857
  • 53,190

2 Answers2

1

Here's how you've defined $(x)$:

$$ (x)= \left\{ (r_0x+xs_0)+(r_1xs_1+\cdots+r_kxs_k)+nx : ~ \begin{array}{ll} r_0,\cdots,r_k \\ s_0,\cdots, s_k \end{array} \in R, ~ n\in\mathbb{Z} \right\}$$

You cannot write something like $r_ix+xr_i+r_ixr_i$ for a generic element of $(x)$ because that forces all of the coefficients to be the same element of $R$, which they needn't be. (Also, not sure what you mean by $\langle r_ixr_i\rangle$ when $r_ixr_i$ is a specific element.)

Note this is all superfluous when $R$ is commutative, because $xs_0=s_0x$ and $r_ixs_i=r_is_ix$ in all of the terms, so really it boils down to $(x)=Rx+\mathbb{Z}x$ when $R$ is commutative.

Let's compare $(x)(y)$ and $(xy)$. Every element of the first looks like a sum of things of the form $(rx+\ell x)(sy+my)$, where $r,s\in R$ and $\ell,m\in\mathbb{Z}$. Every element of the second looks like $(txy+nxy)$, where $t\in R$, $n\in\mathbb{Z}$.

Can you show every element in $(x)(y)$ is in $(xy)$ and vice-versa?

anon
  • 155,259
0

An element in the product $I_xI_y$ is a sum of elements of the form $$ (A_x+B_x+C_x+D_x)(A_y+B_y+C_y+D_y) $$ where $A_x\in\langle RxR\rangle$, $B_x\in Rx$, $C_x\in xR$, $D_x\in \mathbb{Z}x$ and similarly for the other elements.

  • $A_xA_y\in \langle RxR^2yR\rangle\subseteq\langle RxRyR\rangle$
  • $A_xB_y\in\langle RxR^2y\rangle\subseteq RxRy$
  • $A_xC_y\in\langle RxRyR\rangle$
  • $A_xD_y\in\langle RxRy\rangle$
  • $B_xA_y\in\langle RxRyR\rangle$
  • $B_xB_y\in\langle RxRy\rangle$
  • $B_xC_y\in\langle RxyR\rangle$
  • $B_xD_y\in Rxy$
  • $C_xA_y\in\langle xR^2yR\rangle\subseteq\langle xRyR\rangle$
  • $C_xB_y\in\langle xR^2y\rangle\subseteq xRy$
  • $C_xC_y\in\langle xRyR\rangle$
  • $C_xD_y\in xRy$
  • $D_xA_y\in\langle xRyR\rangle$
  • $D_xB_y\in xRy$
  • $D_xC_y\in xyR$
  • $D_xD_y\in\mathbb{Z}xy$

So you get $$ \langle RxRyR\rangle+\langle RxRy\rangle+\langle RxyR\rangle+ \langle xRyR\rangle+xRy+ Rxy+xRy+xyR+\mathbb{Z}xy $$

egreg
  • 244,946